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3 years ago

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An insulated rigid tank is divided into two compartments by a partition. One compartment contains 4 kg of O2 at 20 ºC and 150 kP

a, and the other compartment contains 7 kg of N2 at 40 ºCand 100 kPa. Now the partition is removed, and two gases are allowed to mix.
Determine: (a) the mixture temperature, (b) number of moles of each gas, (c) initial volume occupied by each gas, and (d) final pressure of the mixture. Take the values of specific heat of both gases at 300 K.

2 answers:

valkas [14]3 years ago

5 0

Answer:

Explanation:

Given that,

Mass of oxygen M = 4kg

Temperature of oxygen T.i = 20°C

T.i = 20+273 =293K

Specific heat capacity of oxygen

C = 0.658 J/kgK

And at pressure P = 150Kpa

Mass of Nitrogen m = 7kg

Temperature of Nitrogen t.i = 40°C

t.i = 40+273 = 313K

And at pressure p = 100Kpa

Specific heat capacity of nitrogen

c = 0.743 J/kgK

a. Mixture temperature can be determine from the energy balance of the container

Using Heat energy formula

∆U = 0

For oxygen

H.o = MC∆θ

H.o = MC(T.f - T.i)

H.o = 4×0.658 (T.f - 293)

H.o = 2.632(T.f - 293)

For Nitrogen

H.n = mc∆θ

H.n = mc(T.f - t.i)

H.n = 7×0.743(T.f-313)

H.n = 5.201(T.f-313)

Then,

∆U = 0

H.o + H.n = 0

2.632(T.f - 293) + 5.201(T.f-313) = 0

2.632T.f -771.176+ 5.201T.f-1627.9=0

7.833T.f - 2399.089 = 0

7.833T.f = 2399.089

T.f = 2399.089/7.833

T.f = 306.28 K

b. Number of molecules n?

Number of mole can be calculated by using the formula

n = mass/molar mass

For oxygen,

Molar mass of oxygen O2

O2 = 16×2 = 32g/mol

n = mass of oxygen / molar mass of oxygen

n = 4000/32

n = 125 mole

For Nitrogen

Molar mass of nitrogen N.2

N.2 = 14×2 = 28g/mol

n = mass of nitrogen /molar mass of nitrogen

n' = 7000/28

n' = 250 mole

C. The initial volume of each gas V?

Volume can be calculated using the ideal gas law

PV = nRT

V = nRT/P

R is a constant = 8.314 J/mol•K

For nitrogen

V = n'Rt/p

V = 250 × 8.314 ×313 / 100,000

Vn = 6.51m³

For Oxygen

V = nRT/P

V = 125 × 8.314 × 293 / 150,000

Vo = 2.03m³

d. Final pressure mixture

Molar mass of mixture

M = total mass/total mole

M = (7000+4000)/(250+125)

M = 29.333g/mol

The mixture total number of mole

N = mole of oxygen + mole of nitrogen

N = 250+125 = 375 mol

Then, using ideal gas law

PV = nRT

P = nRT/V

Total mole = 375mol

R = 8.314 J/mol•K

Mixture Temperature T.f = 306.28K

Total Volume Vo+Vn = 2.03+6.51

V = 8.54m³

Then,

P = nRT/V

P = 375×8.314×306.28/8.54

P = 111,815.5 Pa

To K•pa

P = 111.82 K•Pa

o-na [289]3 years ago

4 0

Answer:

Find answers attached

Explanation:

Assuming both gases are idea gas

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Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid,

77julia77 [94]

Answer:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

$L_{BC}^2 = L_{AB}^2 + L_{AC}^2$

$L_{BC}^2 = (3.5+2.5)^2+ 4^2$

$L_{BC}= \sqrt{(6)^2+ 4^2}$

$L_{BC}= \sqrt{36+ 16}$

$L_{BC}= \sqrt{52}$

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The cross -sectional of the cable is calculated by the formula :

$A = \dfrac{\pi}{4}d^2$

where d = 4mm

$A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2$

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection in length $\delta$ ; we can calculate for the force $F_{BC$ by using the formula:

$\delta = \dfrac{F_{BC}L_{BC}}{AE}$

$F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}$

where ;

E = modulus elasticity

$L_{BC}$ = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for $L_{BC}$ and 0.006 m for $\delta$ ; we have:

$F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}$

$F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN$ ---- (1)

Similarly; we can determine the force $F_{BC}$ using the allowable maximum stress; we have the following relation,

$\sigma = \dfrac{F_{BC}}{A}$

${F_{BC}}= {A}*\sigma$

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$\sigma =$ maximum allowable stress

Replacing 190 × 10⁶ Pa for $\sigma$ ; we have :

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Snezhnost [94]

Answer:

Explanation:

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and car with the speed of 180 mph crosses it 36 times

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vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

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here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

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